Problem in integrations

Question

We have to find the integration of $\sqrt{x+\sqrt{x^2+2}}$

I'm not able to start the following integration, as I try to put $x=\sqrt{2}\tan t $. But from this I got no results.

Can anybody please give me a proper start.

Answer 1

Using integration by parts twice:

\begin{align*} I &= \int \sqrt{x+\sqrt{x^2+2}} \,dx \\[4pt] &= x\sqrt{x+\sqrt{x^2+2}}- \int x \, d \left( \sqrt{x+\sqrt{x^2+2}} \right) \\[4pt] &= x\sqrt{x+\sqrt{x^2+2}}- \int \frac{x\left( 1+\frac{x}{\sqrt{x^2+2}} \right)} {2\sqrt{x+\sqrt{x^2+2}}} \, dx \\[4pt] &= x\sqrt{x+\sqrt{x^2+2}}- \int \frac{x\sqrt{x+\sqrt{x^2+2}}} {2\sqrt{x^2+2}} \, dx \\[4pt] &= x\sqrt{x+\sqrt{x^2+2}}-\frac{1}{2} \int \sqrt{x+\sqrt{x^2+2}}\, d\left( \sqrt{x^2+2} \right) \\[4pt] &= x\sqrt{x+\sqrt{x^2+2}}-\frac{\sqrt{x^2+2}}{2} \sqrt{x+\sqrt{x^2+2}} \\ &\quad \: +\frac{1}{2} \int \sqrt{x^2+2} \, d\left( \sqrt{x+\sqrt{x^2+2}} \right) \\[4pt] &= \left( x-\frac{\sqrt{x^2+2}}{2} \right) \sqrt{x+\sqrt{x^2+2}}+ \frac{1}{4} \int \sqrt{x+\sqrt{x^2+2}} \,dx \\[4pt] &= \left( x-\frac{\sqrt{x^2+2}}{2} \right) \sqrt{x+\sqrt{x^2+2}}+ \frac{I}{4}+C' \\[4pt] I &= \frac{2}{3} \left( 2x-\sqrt{x^2+2} \right) \sqrt{x+\sqrt{x^2+2}}+C \end{align*}

Answer 2

Hint: Use Euler substitution: $${x+\sqrt{x^2+2}}=t$$ $$dx=\dfrac{t^2+2}{2t^2}$$ The integral becomes: $$\int\sqrt{t}\times\dfrac{t^2+2}{2t^2}dt=\int\dfrac{\sqrt{t}}{2}+t^{\frac{-3}{2}}dt$$

$$\dfrac{1}{3}\times\dfrac{(t^2-6)}{\sqrt{t}}$$